I participated in a race.1/5th of those who are before
me are equal to 5/6th of those behind me. What were the
total number of contestants in the race?

• If there are x person ahead
  and y persons behind me, then
  total persons are x+y+1
  where
  x/5 = 5y/6
  6x=25 y
  For integral minimum values , x=25 and y=6
  
  then total contestants = 25+6+1=32
• 12 years agoHelpfull: Yes(16) No(3)
• no of contestant before me=x
  no of contestant after me=y
  given:
  x/5=5y/6
  put different values of x & y
  till both the sides become equal
  now when x=25 , y=6
  we will get 5=5
  therefore 25+6+1(me)=32
  
  
• 12 years agoHelpfull: Yes(4) No(0)
• Minimum 32 persons are there.
  
  25 persons are ahead of me and 6 persons are behind me.
  total 25+1+6 = 32 persons.
• 12 years agoHelpfull: Yes(2) No(5)
• Could you please explain it briefly?
  
• 12 years agoHelpfull: Yes(1) No(1)
• Thank you Mr.S.Singh
• 12 years agoHelpfull: Yes(0) No(1)
• Total = x + y + 1
  
  x/5 = 5y/6
  
  x = 25y/6
  
  Total = 31y/6 + 1
  
  If Y = 6, Total = 32
  6 people behind me (y)
  25 people in front of me (x)
  And me.
• 11 years agoHelpfull: Yes(0) No(0)
• let x be the number of participants infront of the person and y be the number of persons behind the person.
  then by the given information in the question
  x/5=(5/6)y
  then x/y=25/6
  therefore x=25 and y=6
  then total number of participants = 25+6+1=32.
• 6 years agoHelpfull: Yes(0) No(0)